+0  
 
0
70
2
avatar

The equation of the line that goes through the point (2,3) and is parallel to the 4x+3y=3 can be written in the form y=mx+b where m is: and b is:

 Dec 12, 2018
 #1
avatar+17163 
+2

Rewrite   4x+3y=3

               3y=-4x+3

                 y = -4/3x +1      where =-4/3 is the slope....parallel line will have same slope

 

y = -4/3 x + b    includes 2,3

3 = -4/3 (2) + b

3+8/3 = b

b = 5 2/3

 

so NEW line is    y = -4/3x + 5 2/3      m = -4/3   b = 5  2/3

 Dec 12, 2018
 #2
avatar+559 
0

We can make \(4x+3y=3\) into \(y=mx+b\) form! We need to isolate \(y\) .

\(3y=-4x+3, y=-\frac{4}{3}x+1\)

However, the line goes through (2, 3) so the y-intercept may not be 1! Let's substitute.

\(3=-\frac{4}{3}\times 2+b, 3=-\frac{8}{3}+b, b=17/3\) and m=-4/3.

 

You are very welcome!

:P

 Dec 12, 2018

7 Online Users